Chapter 3, Problem 60AP

Figure P3.60 illustrates the difference in proportions between the male (m) and female (f) anatomies. The displacements ${\stackrel{\to }{\text{d}}}_{\text{1m}}$. and ${\stackrel{\to }{\text{d}}}_{\text{1f}}$ from the bottom of the feet to the navel have magnitudes of 104 cm and 84.0 cm, respectively. The displacements ${\stackrel{\to }{\text{d}}}_{\text{2m}}$ and ${\stackrel{\to }{\text{d}}}_{\text{2f}}$ have magnitudes of 50.0 cm and 43.0 cm, respectively. (a) Find the vector sum of the displacements ${\stackrel{\to }{\text{d}}}_{\text{d}1}$ and ${\stackrel{\to }{\text{d}}}_{\text{d2}}$ in each case. (b) The male figure is 180 cm tall, the female 168 cm. Normalize the displacements of each figure to a common height of 200 cm and re-form the vector sums as in part (a). Then find the vector difference between the two sums.

  

  Figure P3.60

To determine

The vector sum of the displacements in each cases.

Answer to Problem 60AP

Solution:

The magnitude of the vector sum of displacements are $132\text{cm}$ and $111\text{cm}$, the direction of these vectors respectively $69.7°$ and $69.9°$ respectively.

Explanation of Solution

Given Info:

The x-components of $d_{1m}$ is zero, y- component of $d_{1m}$ is $104\text{cm}$, the x-component of $d_{1f}$ is zero, the y-component of $d_{1f}$ is $84.0\text{cm}$, the angle made by $d_{2m}$ with horizontal is $23.0°$, the angle made by $d_{2f}$ with respect to horizontal is $28.0°$, the magnitude of $d_{2m}$ is $50.0\text{cm}$ and the magnitude of $d_{2f}$ is $43.0\text{cm}$.

Write the formula to calculate the x-component of $d_{2m}$.

$x_{2m}=d_{2m}\cos {\theta }_1$

• $x_{2m}$ is the x-component of $d_{2m}$
• ${\theta }_1$ is the angle made by $d_{2m}$ with respect to horizontal

Substitute $50.0\text{cm}$ for $d_{2m}$ and $23.0°$ for ${\theta }_1$ to calculate $x_{2m}$.

$\begin{array}{c}{x}_{2m}=\left(50.0\text{cm}\right)\cos 23.0°\\ =\left(50.0\text{cm}\right)\left(0.92\right)\\ =46\text{cm}\end{array}$

Write the formula to calculate the y-component of $d_{2m}$.

$y_{2m}=d_{2m}\sin {\theta }_1$

Substitute $50.0\text{cm}$ for $d_{2m}$ and $23.0°$ for ${\theta }_1$ to calculate $y_{2m}$.

$\begin{array}{c}{y}_{2m}=\left(50.0\text{cm}\right)\sin 23.0°\\ =\left(50.0\text{cm}\right)\left(0.39\right)\\ =19.5\text{cm}\end{array}$

Write the formula to calculate the x-component of $d_{2f}$.

$x_{2f}=d_{2f}\cos {\theta }_2$

• $x_{2f}$ is the x-component of $d_{2f}$
• ${\theta }_2$ is the angle made by $d_{2f}$ with respect to horizontal

Substitute $43.0\text{cm}$ for $d_{2f}$ and $28.0°$ for ${\theta }_2$ to calculate $x_{2f}$.

$\begin{array}{c}{x}_{2f}=\left(43.0\text{cm}\right)\cos 28.0°\\ =\left(43.0\text{cm}\right)\left(0.88\right)\\ =38\text{cm}\end{array}$

Write the formula to calculate the y-component of $d_{2f}$.

$y_{2f}=d_{2f}\sin {\theta }_2$

Substitute $43.0\text{cm}$ for $d_{2f}$ and $28.0°$ for ${\theta }_2$ to calculate $y_{2f}$.

$\begin{array}{c}{y}_{2f}=\left(43.0\text{cm}\right)\sin 28.0°\\ =\left(43.0\text{cm}\right)\left(0.47\right)\\ =20.2\text{cm}\end{array}$

Write the formula to calculate the vector sum of displacement of the vectors.

$d_{d1}=\sqrt{{\left(x_{1m}+x_{2m}\right)}^2+{\left(y_{1m}+y_{2m}\right)}^2}$ (I)

• $d_{d1}$ is the magnitude of the vector $d_{d1}$.

$d_{d2}=\sqrt{{\left(x_{1f}+x_{2f}\right)}^2+{\left(y_{1f}+y_{2f}\right)}^2}$ (II)

• $d_{d2}$ is the magnitude of the vector $d_{d2}$.
• $x_{1m}$ is the x-component of the vector $d_{1m}$
• $y_{1m}$ is the y-component of the vector $d_{1m}$
• $x_{1f}$ is the x-component of the vector $d_{1f}$
• $y_{1f}$ is the y-component of the vector $d_{1f}$

Substitute 0 for $x_{1m}$, $46\text{cm}$ for $x_{2m}$, $104\text{cm}$ for $y_{1m}$, and $19.5\text{cm}$ for $y_{2m}$ in (I) to calculate $d_{d1}$.

$\begin{array}{c}{d}_{d1}=\sqrt{{\left(0+46\text{cm}\right)}^2+{\left(104\text{cm}+19.5\text{cm}\right)}^2}\\ =\sqrt{{\left(46\text{cm}\right)}^2+{\left(123.5\text{cm}\right)}^2}\\ =\sqrt{2116{\text{cm}}^{\text{2}}+15252.3{\text{cm}}^{\text{2}}}\\ =132\text{cm}\end{array}$

Substitute 0 for $x_{1f}$, $38\text{cm}$ for $x_{2f}$, $84.0\text{cm}$ for $y_{1f}$ and $20.2\text{cm}$ for $y_{2f}$ in (II) to calculate

$d_{d2}$.

$\begin{array}{c}{d}_{d2}=\sqrt{{\left(0+38\text{cm}\right)}^2+{\left(84\text{cm}+20.2\text{cm}\right)}^2}\\ =\sqrt{{\left(38\text{cm}\right)}^2+{\left(104.2\text{cm}\right)}^2}\\ =\sqrt{1444{\text{cm}}^{\text{2}}+10857.6{\text{cm}}^{\text{2}}}\\ =\sqrt{12301.6{\text{cm}}^{\text{2}}}\\ =111\text{cm}\end{array}$

Write the formula to calculate the direction of two vectors.

${\theta }_m={\tan }^{-1}\left(\frac{y_{1m}+y_{2m}}{x_{1m}+x_{2m}}\right)$ (III)

• ${\theta }_m$ is the direction of the vector $d_{d1}$

${\theta }_f={\tan }^{-1}\left(\frac{y_{1f}+y_{2f}}{x_{1f}+x_{2f}}\right)$ (IV)

• ${\theta }_f$ is the direction of the vector $d_{d2}$

Substitute 0 for $x_{1m}$, $46\text{cm}$ for $x_{2m}$, $104\text{cm}$ for $y_{1m}$, and $19.5\text{cm}$ for $y_{2m}$ in (III) to calculate

${\theta }_m$.

$\begin{array}{c}{\theta }_m={\tan }^{-1}\left(\frac{104\text{cm}+19.5\text{cm}}{0+46\text{cm}}\right)\\ ={\tan }^{-1}\left(\frac{104\text{cm}+19.5\text{cm}}{0+46\text{cm}}\right)\\ ={\tan }^{-1}\left(\frac{123.5\text{cm}}{46\text{cm}}\right)\\ ={\tan }^{-1}\left(2.7\right)\\ =69.7°\end{array}$

Substitute 0 for $x_{1f}$, $38\text{cm}$ for $x_{2f}$, $84.0\text{cm}$ for $y_{1f}$ and $20.2\text{cm}$ for $y_{2f}$ in (IV) to calculate

${\theta }_f$.

$\begin{array}{c}{\theta }_f={\tan }^{-1}\left(\frac{84.0\text{cm}+20.2\text{cm}}{0+38\text{cm}}\right)\\ ={\tan }^{-1}\left(\frac{104.2\text{cm}}{38\text{cm}}\right)\\ ={\tan }^{-1}\left(\frac{123.5\text{cm}}{46\text{cm}}\right)\\ ={\tan }^{-1}\left(2.74\right)\\ =69.9°\end{array}$

Thus, the direction of these vectors respectively $69.7°$ and $69.9°$.

Conclusion:

Therefore, the magnitude of the vector sum of displacements are $132\text{cm}$ and $111\text{cm}$, the direction of these vectors respectively $69.7°$ and $69.9°$ respectively.

To determine

The normalized vectors and magnitude.

Answer to Problem 60AP

Solution:

The magnitude of normalized vector sum of displacements are $146\text{cm}$ and $132\text{cm}$, the difference in the normalized vector is $14\text{cm}$ and direction of the normalized vectors is $66°$

Explanation of Solution

Given Info:

Common height is $200\text{cm}$, height of the male is $180\text{cm}$ and height of the female is $168\text{cm}$.

Write the formula to calculate scale factor for $d_{d1}$.

$S_m=\frac{H}{h_m}$

• $S_m$ is the scale factor for $d_{d1}$
• H is the common height
• $h_m$ is the height of the male

Substitute $200\text{cm}$ for H and $180\text{cm}$ for $h_m$ to calculate $S_m$.

$\begin{array}{c}{S}_m=\frac{200\text{cm}}{180\text{cm}}\\ =1.11\end{array}$

Write the formula to calculate scale factor for and $d_{d2}$.

$S_f=\frac{H}{h_f}$

• $S_m$ is the scale factor for $d_{d1}$
• H is the common height
• $h_f$ is the height of the female

Substitute $200\text{cm}$ for H and $168\text{cm}$ for $h_m$ to calculate $S_f$.

$\begin{array}{c}{S}_f=\frac{200\text{cm}}{168\text{cm}}\\ =1.19\end{array}$

Write the formula to calculate the normalized vector sum of $d_{d1}$

$d_m{}^{\prime }=S_m{d}_{d1}$

• $d_m{}^{\prime }$ is the normalized vector of $d_{d1}$

Substitute $132\text{cm}$ for $d_{d1}$ to calculate $d_m{}^{\prime }$.

$\begin{array}{c}{d}_m{}^{\prime }=\left(1.11\right)\left(132\text{cm}\right)\\ =146\text{cm}\end{array}$

Write the formula to calculate the normalized vector sum of $d_{d2}$

$d_f{}^{\prime }=S_f{d}_{d2}$

• $d_f{}^{\prime }$ is the normalized vector of $d_{d2}$

Substitute $111\text{cm}$ for $d_{d1}$ to calculate $d_f{}^{\prime }$.

$\begin{array}{c}{d}_f{}^{\prime }=\left(1.19\right)\left(111\text{cm}\right)\\ =132\text{cm}\end{array}$

Write the formula to calculate the normalized x-component for $d_{d1}$.

$x_m{}^{\prime }=\left(x_{1m}+x_{2m}\right)s_m$

• $x_m{}^{\prime }$ is the normalized x-component for $d_{d1}$

Substitute 0 for $x_{1m}$, $46.0\text{cm}$ for $x_{2m}$ and $1.11$ for $S_m$ to calculate $x_m{}^{\prime }$.

$\begin{array}{c}{x}_m{}^{\prime }=\left(0+46.0\text{cm}\right)1.11\\ =\left(46.0\text{cm}\right)1.11\\ =51\text{cm}\end{array}$

Write the formula to calculate the normalized y-component for $d_{d1}$.

$y_m{}^{\prime }=\left(y_{1m}+y_{2m}\right)s_m$

• $y_m{}^{\prime }$ is the normalized y-component for $d_{d1}$

Substitute $104\text{cm}$ for $y_{1m}$, $19.5\text{cm}$ for $y_{2m}$ and $1.11$ for $S_m$ to calculate $y_m{}^{\prime }$.

$\begin{array}{c}{y}_m{}^{\prime }=\left(104\text{cm}+19.5\text{cm}\right)1.11\\ =\left(123.5\text{cm}\right)1.11\\ =137\text{cm}\end{array}$

Write the formula to calculate the normalized x-component for $d_{d2}$.

$x_f{}^{\prime }=\left(x_{1f}+x_{2f}\right)s_f$

• $x_f{}^{\prime }$ is the normalized x-component for $d_{d2}$

Substitute 0 for $x_{1f}$, $38.0\text{cm}$ for $x_{2f}$ and $1.19$ for $S_f$ to calculate $x_f{}^{\prime }$.

$\begin{array}{c}{x}_f{}^{\prime }=\left(0+38.0\text{cm}\right)1.19\\ =\left(38.0\text{cm}\right)1.19\\ =45\text{cm}\end{array}$

Write the formula to calculate the normalized y-component for $d_{d2}$.

$y_f{}^{\prime }=\left(y_{1f}+y_{2f}\right)s_f$

• $y_f{}^{\prime }$ is the normalized y-component for $d_{d2}$

Substitute $84.0\text{cm}$ for $y_{1f}$, $20.2\text{cm}$ for $y_{2f}$ and $1.19$ for $S_f$ to calculate $y_f{}^{\prime }$.

$\begin{array}{c}{y}_f{}^{\prime }=\left(84.0\text{cm}\text{cm}+20.2\text{cm}\right)1.19\\ =\left(104.2\text{cm}\right)1.19\\ =124\text{cm}\end{array}$

Write the formula to calculate the difference in the normalized vector.

$d^{\prime }=\sqrt{{\left(x_m{}^{\prime }-x_f{}^{\prime }\right)}^2+{\left(y_m{}^{\prime }-y_f{}^{\prime }\right)}^2}$

• $d^{\prime }$ is the difference in the normalized vectors

Substitute $51\text{cm}$ for $x_m{}^{\prime }$, $45\text{cm}$ for $x_f{}^{\prime }$, $137\text{cm}$ for $y_m{}^{\prime }$ and $124\text{cm}$ for $y_f{}^{\prime }$ to calculate $d^{\prime }$.

$\begin{array}{c}{d}^{\prime }=\sqrt{{\left(51\text{cm}-45\text{cm}\right)}^2+{\left(137\text{cm}-124\text{cm}\right)}^2}\\ =\sqrt{{\left(6\text{cm}\right)}^2+{\left(13\text{cm}\right)}^2}\\ =\sqrt{205{\text{cm}}^2}\\ =14\text{cm}\end{array}$

Write the formula to calculate the direction of the normalized vector.

$\theta \text{'}={\tan }^{-1}\left(\frac{y_m{}^{\prime }-y_f{}^{\prime }}{x_m{}^{\prime }-x_f{}^{\prime }}\right)$

• ${\theta }^{\prime }$ is the in the direction of the normalized vector

Substitute $51\text{cm}$ for $x_m{}^{\prime }$, $45\text{cm}$ for $x_f{}^{\prime }$, $137\text{cm}$ for $y_m{}^{\prime }$ and $124\text{cm}$ for $y_f{}^{\prime }$ to calculate ${\theta }^{\prime }$.

$\begin{array}{c}\theta \text{'}={\tan }^{-1}\left(\frac{137\text{cm}-124\text{cm}}{51\text{cm}-45\text{cm}}\right)\\ ={\tan }^{-1}\left(\frac{13\text{cm}}{6\text{cm}}\right)\\ ={\tan }^{-1}\left(2.2\right)\\ =66°\end{array}$

Conclusion:

Therefore, the magnitude of normalized vector sum of displacements are $146\text{cm}$ and $132\text{cm}$, the difference in the normalized vector is $14\text{cm}$ and direction of the normalized vectors is $66°$